It seems that from this webpage, the spin cobordism is equivalent to KO theory in low dimension.

If we denote the $p$-torsion part (mean $\mathbb{Z}_{p^n}$ for some $n$) $$\Omega_d(BG)_p.$$

Question 1: Then do we have $$\Omega_d^{spin}(BG)_p = ko_d(BG)_p?$$ for $p=2$ and free part, for $d\le 7$? (how about higher $d>7$?)

And $$ \Omega_d^{spin}(BG)_p = \Omega_d^{SO}(BG)? $$ for $p \neq 2$ and $p$ is an odd prime?

Namely, the 2-torsion and free part of $Mspin$ and $KO$ is the same. If there is an odd $p$ torsion, we need to consider localization at odd prime by $MSO$ cohomology. Is this correct?

Question 2: If this is a statement about the spectra, not just about stable homotopy groups, and thus within these spin cobordism and ko theory, do they completely coincide for any dimensions $d$, instead of just $d \leq 7$?

  • i) By ko, do you mean the connective real K-theory? ii) What is G here? A finite group? Discrete greoup? Compact Lie group? Any topological group? – user43326 Oct 16 at 4:54
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    @user43326 in the context wonderich is asking about, $\mathit{ko}$ is indeed connective real $K$-theory, and I believe $G$ can be any compact Lie group. – Arun Debray Oct 22 at 4:34

First of all, let me say that the page you are quoting is a little bit misleading if not inaccurate on the Anderson-Brown-Peterson splitting.

$ko\langle 4n(J)\rangle $ should read $\Sigma ^{4n(J)} ko$ and $ko\langle 4n(J)-2\rangle $ should read $\Sigma ^{4n(J)-4} ko\langle 2\rangle $

See, e.g. https://pdfs.semanticscholar.org/c377/7dd83a6ba0d959c5c62d633ed4109cddb660.pdf

With this correction, at 2, $Mspin \wedge BG$ splits whose bottom piece is $ko\wedge BG$, other pieces are at least 7-connected since the "next bottom" piese is $\Sigma ^8ko \wedge BG$ as is pointed out by ArunDebray, corresponding to the partition $J=(2)$.

Thus the answer to your questions, at the prime 2 is that

  1. We have an isomorphism up to $d\leq 7$
  2. The map is always surjective, but the kernel is in general non-trivial for $d\geq 8$.
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    Are you sure it's a $\Sigma^4\mathit{ko}$, and not a $\Sigma^8\mathit{ko}$? $\pi_5\mathit{MSpin}$ and $\pi_6\mathit{MSpin}$ both vanish, but if $\mathit{MSpin}$ had a $\Sigma^4\mathit{ko}$ summand, they would both contain a $\mathbb Z/2$ summand. – Arun Debray Oct 22 at 16:51
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    @ArunDebray You are right, $n(J)$ even, so we get $\Sigma ^8ko$. With $n(J)$ odd sequences, we are not allowed to have 1 so the lowest is $\Sigma 8ko<2>$. Presumably in the range the op asks, there is no HZ/2 summand, I will correct my answer later, thanks. – user43326 Oct 23 at 8:31

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